For what values of x is #f(x)= x^3 - 4x^2 + 5x # concave or convex?

Question

Emily Ammerman · Accepted Answer

concave #(-oo,4/3)#
convex#(4/3,+oo)# To determine where f(x) is concave/convex we require to find f''(x) f(x)#=x^3-4x^2+5x# f'(x)#=3x^2-8x+5# and f''(x)#=6x-8# We now equate f''(x) to zero to find values of x where any change from concave/convex or convex/concave may occur. solve : 6x - 8 = 0 #rArr x=4/3# We now have to check the value of f''(x) to the left and right of #x=4/3 , "say " x=a# • If f''(a) > 0 , then f(x) is convex • If f''(a) < 0 , then f(x) is concave x = 0 is to the left and f''(0) = - 8 → concave x = 2 is to the right and f''(2) = 4 → convex #"hence" f(x)" is concave " (-oo,4/3)# and f(x)#" is convex " (4/3,+oo)# graph{x^3-4x^2+5x [-10, 10, -5, 5]}

Samantha Adkison · Answer

undefined To determine where the function $ f(x) = x^3 - 4x^2 + 5x $ is concave or convex, we need to find the second derivative of the function. The second derivative of $ f(x) $ is given by: $$ f''(x) = \frac{d^2}{dx^2} (x^3 - 4x^2 + 5x) $$ $$ = \frac{d}{dx} (3x^2 - 8x + 5) $$ $$ = 6x - 8 $$ For concavity, if $ f''(x) > 0 $, the function is concave up, and if $ f''(x) < 0 $, the function is concave down. For convexity, if $ f''(x) > 0 $, the function is convex up, and if $ f''(x) < 0 $, the function is convex down. Setting $ f''(x) = 0 $ gives us the critical point: $$ 6x - 8 = 0 $$ $$ 6x = 8 $$ $$ x = \frac{8}{6} = \frac{4}{3} $$ Now, we can test intervals around the critical point to determine the concavity or convexity of the function. For $ x < \frac{4}{3} $: $$ f''(x) = (6x - 8) < 0 $$ So, the function is concave down. For $ x > \frac{4}{3} $: $$ f''(x) = (6x - 8) > 0 $$ So, the function is concave up. Therefore, the function $ f(x) = x^3 - 4x^2 + 5x $ is concave for $ x < \frac{4}{3} $ and convex for $ x > \frac{4}{3} $.